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Thread: Points on graphs

  1. #1
    Junior Member Fnus's Avatar
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    Points on graphs

    The two exercises I'd like help with is on the picture, thanks (:
    Attached Thumbnails Attached Thumbnails Points on graphs-1.jpg  
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  2. #2
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    Hello, Fnus!

    Here's the first one . . .


    $\displaystyle f(x) \:=\:x^3 + ax + b$ has a stationary point at $\displaystyle (-2,3).$

    (a) Find the values of $\displaystyle a$ and $\displaystyle b.$

    (b) Find the position and nature of all stationary points.
    The point (-2,3) is on the graph.
    . . Hence: .$\displaystyle f(\text{-}2) = 3\!:\;\;(\text{-}2)^3 + a(\text{-}2) + b \:=\:3\quad\Rightarrow\quad \text{-}2a + b \:=\:11$ .[1]

    $\displaystyle f'(x) \:=\:3x^2 + a$
    Since (-2,3) is a stationary point, $\displaystyle f'(\text{-}2) = 0$
    . . and we have: .$\displaystyle 3(\text{-}2)^2 + a \:=\:0\quad\Rightarrow\quad\boxed{a \:=\:\text{-}12}$

    Substitute into [1]: .$\displaystyle \text{-}2(\text{-}12) + b \:=\:11\quad\Rightarrow\quad\boxed{b \:=\:\text{-}13}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The function is: .$\displaystyle f(x) \:=\:x^3 - 12x - 13$

    Then: .$\displaystyle f'(x) = 0\quad\Rightarrow\quad 3x^2 - 12 \:=\:0\quad\Rightarrow\quad x^2\:=\:4\quad\Rightarrow\quad x \:=\:\pm2$

    When $\displaystyle x = 2\!:\;\;f(2) \:=\:2^3 - 12(2) - 13 \:=\:-29$ . . . Critical point: $\displaystyle (2,\,\text{-}29)$

    When $\displaystyle x = \text{-}2\!:\;\;f(\text{-}2) \:=\:(\text{-}2)^3 - 12(\text{-}2) - 13 \:=\:3$ . . . Critical point: $\displaystyle (\text{-}2,\,3)$


    Second Derivative Test: .$\displaystyle f''(x) \:=\:6x$

    $\displaystyle f''(2) \:=\:6(2) \:=\:+12$ . . . concave up . . . minumum at $\displaystyle {\color{blue}(2, \text{-}29)}$

    $\displaystyle f''\text{(}-2) \:=\:6(\text{-}2) \:=\:-12$ . . . concave down . . . maximum at $\displaystyle {\color{blue}(\text{-}2, 3)}$

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  3. #3
    Junior Member Fnus's Avatar
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    Oh, I see.
    Thanks (:
    That step by step thing makes me actually understand what is going on.
    Thanks a lot!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Here's 6

    Let $\displaystyle P(x) = ax^3 + bx^2 + cx + d$

    $\displaystyle \Rightarrow P'(x) = 3ax^2 + 2bx + c$

    Now, let's piece our clues together.

    Clue 1: $\displaystyle P(x)$ touches the line $\displaystyle y = 9x + 2$ at the point $\displaystyle (0,2)$

    we can take two things away from this.

    (1) we interpret the phrase "touching the line" to mean, the line is tangent to the curve. thus it means that $\displaystyle P'(0) = 2 \implies \boxed{c = 2}$

    (2) the second thing we can take from this, is that $\displaystyle P(x)$ passes through the point $\displaystyle (0,2)$, that is $\displaystyle P(0) = 2$ which means of course, $\displaystyle \boxed{d = 2}$

    two unknowns down, two to go.

    what else do we know about $\displaystyle P(x)$?

    Clue 2: $\displaystyle P(x)$ has a stationary point at $\displaystyle (-1,-7)$

    again, there are two things we can take from this.

    (1) $\displaystyle P(x)$ passes through $\displaystyle (-1,-7)$, that is, $\displaystyle P(-1) = -7$

    that means that: $\displaystyle a(-1)^3 + b(-1)^2 - 2 + 2 = -7$ (i plugged in the values we got for c and d)

    simplifying, we get:

    $\displaystyle -a + b = -7$ ......................(1)

    (2) the second thing we take from this is that $\displaystyle P'(-1) = 0$ (that is, the slope of $\displaystyle P(x)$ when $\displaystyle x = -1$ is zero)

    but that means that: $\displaystyle 3a(-1)^2 - 2b - 2 = 0$ (i plugged in the value for c)

    simplifying we get:

    $\displaystyle 3a - 2b = -2$ ......................(2)


    thus, to find the remaining unknowns, we must solve the system

    $\displaystyle -a + b = -7$ ......................(1)
    $\displaystyle 3a - 2b = -2$ ......................(2)

    the solutions are $\displaystyle \boxed{a = -16}$ and $\displaystyle \boxed{b = -23}$

    thus, $\displaystyle P(x) = -16x^3 - 23x^2 + 2x + 2$



    EDIT: i made an error. when i concluded that c = 2, i claimed the wrong thing, i should have said that $\displaystyle P'(0) = 9 \implies \boxed{c = 9}$ (since 9 is the slope of the line). make the necessary corrections, i can't bother
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